AP Calculus AB and BC

Limit of arcsin 2x / x at 0 Is 2

Q: What is the general rule for arcsin(ax) over x?

It tends to a as x \to 0. The same holds with \arctan in place of \arcsin.

The limit of arcsin 2x over x as x approaches 0 is 2. Near 0 the arcsine of a small number is almost that number, so arcsin 2x behaves like 2x and the quotient behaves like 2x over x. Matching the denominator to the inner angle finishes it in one line.

\lim_{x \to 0} \frac{\arcsin 2x}{x} = 2

Settled by matching the inner angle to the denominator.

Put the inner angle under the numerator

The standard limit is $\lim_{u \to 0}\frac{\arcsin u}{u} = 1$ , so the denominator has to be $2x$ , not $x$ . Multiply and divide by $2$ to arrange that.

\frac{\arcsin 2x}{x} = 2\cdot\frac{\arcsin 2x}{2x}

As $x \to 0$ the inner quantity $u = 2x$ tends to $0$ as well, so the quotient tends to $1$ and the constant $2$ is all that remains.

\lim_{x \to 0}\frac{\arcsin 2x}{x} = 2\lim_{u \to 0}\frac{\arcsin u}{u} = 2

Why arcsine behaves like its input

Set $u = \arcsin t$ , so that $t = \sin u$ and $u \to 0$ as $t \to 0$ . The unfamiliar quotient turns into a familiar one.

\frac{\arcsin t}{t} = \frac{u}{\sin u} \longrightarrow 1

The graphs say the same thing. Sine and arcsine are reflections of each other in the line $y = x$ , and sine leaves the origin with slope $1$ , so its reflection leaves the origin with slope $1$ too.

One family, one trick

Any function passing through the origin with slope 1 behaves like its input near 0. That covers sin, tan, arcsin, arctan and the natural log of 1 plus u. A ratio of two of them, with linear inner angles, tends to the ratio of the coefficients.

The mistakes students make

Notation and coefficients cause the trouble here, not the limit itself.

Answering $1$ by treating the expression as the standard limit as it stands. The inner angle is $2x$ while the denominator is $x$ , so a factor of $2$ is missing.
Answering $\frac{1}{2}$ by dividing by the coefficient instead of multiplying by it.
Reading $\sin^{-1}2x$ as $\frac{1}{\sin 2x}$ , which would send the quotient to $\infty$ . The $-1$ marks an inverse function, not a reciprocal.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

What is the limit of arcsin 2x / x as x approaches 0?

It is $2$ .

Why does arcsin behave like sin near 0?

They are inverse functions, so their graphs are mirror images in the line $y = x$ . Sine has slope $1$ at the origin, and reflecting a slope of $1$ leaves it unchanged.

What is the general rule for arcsin(ax) over x?

It tends to $a$ as $x \to 0$ . The same holds with $\arctan$ in place of $\arcsin$ .